189. Rotate Array

• Link: https://leetcode.com/problems/rotate-array/description/

Question

Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation:
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

Follow up:

• Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

Approach 1: Brute Force

Algorithm

1. Create a copy of the nums vector nums2
2. Iterate through nums with i using for loop
3. Replace nums[(i+k)%n] with nums2[i]

Code

class Solution {
public:
	void rotate(vector<int>& nums, int k) {
		vector<int> nums2(nums);
		int n = nums.size();
		
		for (int i = 0 ; i < n ; i++) {
			nums[(i+k) % n] = nums2[i];
		}
		return;
	}
};

Complexity Analysis

• Time Complexity: $O(n)$ (1 loop through entire array)
• Space Complexity: $O(n)$ (new array of size n)

Approach 2: Optimal

Algorithm: Using Reversal Algorithm

1. Reverse function
   1. Arguments: nums(passed by reference), start, end index
   2. while start <= end
      1. temp = nums[start];
      2. nums[start] = nums[end];
      3. nums[end] = temp;
      4. start++, end--
2. Reverse first n-k elements reverse(nums, 0, n - k - 1)
3. Reverse last k elements reverse(nums, n-k, n-1)
4. Reverse the while array reverse(nums, 0, n-1)

Code

class Solution {
public:
	void reverse(vector<int>& nums, int start, int end) {
		while (start < end) {
			int temp = nums[start];
			nums[start] = nums[end];
			nums[end] = temp;
			start++;
			end--;
		}
	}
	
	void rotate(vector<int>& nums, int k) {
		int n = nums.size();
		// Normalize k
		k = k % n;
		
		// Reverse the first part (0 to n-k-1)
		reverse(nums, 0, n - k - 1);
		
		// Reverse the second part (n-k to n-1)
		reverse(nums, n - k, n - 1);
		
		// Reverse the whole array (0 to n-1)
		reverse(nums, 0, n - 1);
	}
};

Complexity Analysis

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$ (no extra memory used)